Question

one root of the equation 2x^6-3x^5+5x^4+6x^3-27x+81=0, is √2+¡, find the remaining roots.

Answer 1

Answer:

The roots of the equation are $\sqrt{2}+i ,\sqrt{2}-i ,3 ,3/2 ,7-4\sqrt{2}i ,-7-4\sqrt{2}i$

Step-by-step explanation:

Given,

$2x^6-3x^5+5x^4+6x^3-27x+81=0$

One of the roots of the equation = $\sqrt{2}+i$

We need to find the rest of the roots.

Since one of the root is a complex number then it's conjugate will also be a root of the equation.$\sqrt{2}-i$

By separating the equation we can write the equation as,

$2x^6 - 3x^5 + 5x^4 + 6x^2 - 27x + 81 = (2x^2 - 3x + 9)(x^4 - 2x^2 + 9)$ since,

$(x^2 - 2(2)x + 3)(x^2 + 2(2)x + 3) = x^4 - 2x^2 + 9$

Now we can simply factorise both terms as two separate equations.

$2x^2-3x+9=0$

$2x^2-6x+3x+9=0$

$x = 3$ and $x = 3/2$

Taking the other equation,

we first substitute $y=x^2$, We get an equation.

$y^2-2x+9=0$

By applying,

$(-b+\sqrt{b^2-4ac})/2$ and $(-b-\sqrt{b^2-4ac})/2$

we get,

y = $-1-2\sqrt{2}i$ and y = $-1+2\sqrt{2}i$, Hence

x = $7 - 4\sqrt{2} i$ and x = $-7-4\sqrt{2}i$

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